Chemistry Examination Discussion

[andyse7en17]

andyse7en17 - check my previous post.

Just did. That's what I thought. I actually put reasonings for both answers and indicate that "below is my explanation"... do you think I'll still get penalized?

[Emily C]

and we added exactly half the amount of OH- to neutralise the CH3COOH, we are effectively removing reactant and adding product, and their concentrations will be equal at this point. And you are right by saying that by Le Chatelier's Principle, some of the CH3COO- will react with the H+ to form CH3COOH subsequently. So technically, [CH3COOH] will be ever so slightly higher than [CH3COO-] and hence Ka > [H+].

Thats how i interpreted it because its a weak acid so concentration of ethonic acid is always be slightly more than ethanoate ions, not a good question, could have been interpreted in different ways

[Edward21]

Gah FML. I had 80% purity, thought that for 3 marks this was way too easy, did something else with an elaborate explanation and got 11.6% purity.. 
Could someone tell me what the 3 marks would be for in the question? I may be able to get 2/3 with the right working for everything else.
I reasoned that if you electrolysed copper anode, pure copper cathode in the Cu solution, the copper concentration shouldn't change, but rather other ions like Co2+ and Ni2+ would be oxidised from their metals preferentially to the Cu; but Cu would still be reduced from the solution of the cathode in this time so you could get reduced Cu at the cathode, whilst ions are forming that aren't Cu2+ at the anode. I was thinking about this question like that weak acid question last year with only 1% of the state getting it right. I thought more complex thought processes were needed for this question, obviously not  I can't believe I had the right answer straight up 

[andyse7en17]

and we added exactly half the amount of OH- to neutralise the CH3COOH, we are effectively removing reactant and adding product, and their concentrations will be equal at this point. And you are right by saying that by Le Chatelier's Principle, some of the CH3COO- will react with the H+ to form CH3COOH subsequently. So technically, [CH3COOH] will be ever so slightly higher than [CH3COO-] and hence Ka > [H+].

It's the opposite. Some of the CH3COOH will reacts to form CH3COO- since initially Q=(1 x 10^-1)/1 < Ka (assuming initially [CH3COOH]=[CH3COO-]=1M)... But yeah, [H+]<Ka

[lzxnl]

They're asking for which one is the best choice, and you probably wouldn't be able to measure the difference, so saying they're equal is probably best.

[simba]

It's the opposite. Some of the CH3COOH will reacts to form CH3COO- since initially Q=(1 x 10^-1)/1 < Ka (assuming initially [CH3COOH]=[CH3COO-]=1M)... But yeah, [H+]<Ka

You've got guts to argue with the chemistry god

[Stick]

FML I ticked the first box thinking that water would drive the ionisation forwards a bit more. I think I underestimated this exam. XD

For the sake of me learning after the exam, could someone explain why my reasoning was wrong for this? XD

[andyse7en17]

They're asking for which one is the best choice, and you probably wouldn't be able to measure the difference, so saying they're equal is probably best.

that's what I got confused. "best choice"  I was able to measure the difference!!! I wrote my explanation and calculation down.

[joey7]

It's the opposite. Some of the CH3COOH will reacts to form CH3COO- since initially Q=(1 x 10^-1)/1 < Ka (assuming initially [CH3COOH]=[CH3COO-]=1M)... But yeah, [H+]<Ka

I agree with this guy, not that my opinion means much
No wait I think I understand Thushans' explanation now

[thushan]

It's the opposite. Some of the CH3COOH will reacts to form CH3COO- since initially Q=(1 x 10^-1)/1 < Ka (assuming initially [CH3COOH]=[CH3COO-]=1M)... But yeah, [H+]<Ka

Nah, Q is greater than Ka initially - Ka = 1.7 x 10^-5, and your initial [H+] is 1.8 x 10^-3.

[thushan]

What a gay question

Haha. I think VCAA set it as a question to get people to think, because this is something commonly taught in university, but I think they didn't realise that Year 12 students who are worried about getting "one" correct answer will be confused as to what VCAA wanted.

[presto]

Ok, I'm not sure if my reasoning was completely right and chem's definitely not my forte, but this is how I thought about it:
There's if half the OH- is added, then:
the ratio of CH3COOH:CH3COO- = 1:1
but there is that ionisation reaction of CH3COOH in water as well. Here I noticed the Ka for the CH3COOH was ~10^-5
and I recalled something about how if it's below 10^-2, then there are generally more reactants than products.
So I arrived at the conclusion: [CH3COOH]>[CH3COO-].
But... I read the options wrong and ticked the 1st box...

As for the next bit, I decided to keep it simple (as I was not 100% sure about the previous part) and said:
Ka=[H3O+].

[charmanderp]

I think everyone saying that this exam was straightforward has underestimated it a bit - certainly not impossible, but I think there's enough challenging material in there. Never underestimate the amount of detail they require for written answers, too.

[thushan]

Ok, I'm not sure if my reasoning was completely right and chem's definitely not my forte, but this is how I thought about it:
There's if half the OH- is added, then:
the ratio of CH3COOH:CH3COO- = 1:1
but there is that ionisation reaction of CH3COOH in water as well. Here I noticed the Ka for the CH3COOH was ~10^-5
and I recalled something about how if it's below 10^-2, then there are generally more reactants than products.
So I arrived at the conclusion: [CH3COOH]>[CH3COO-].
But... I read the options wrong and ticked the 1st box...

As for the next bit, I decided to keep it simple (as I was not 100% sure about the previous part) and said:
Ka=[H3O+].

Hey presto (hehe nice pun) - "if it's below 10^-2, then there are generally more reactants than products" - that's not something to be religiously followed, you could have relatively more product if the concentration if one of your products is excruciatingly small compared to that of the other products and reactants.

[andyse7en17]

Nah, Q is greater than Ka initially - Ka = 1.7 x 10^-5, and your initial [H+] is 1.8 x 10^-3.

You meant before we put NaOH in? Now I'm even more confused... Does NaOH react with H+ first?